P1196 [NOI2002]银河英雄传说 【带权并查集】

思路

　　用sum记录每个舰队的战舰数量， tohead 记录当前舰离舰首的距离，那么求任意两舰之间有多少舰显然就是 abs( tohead[i] - tohead[j] ) - 1；

CODE

 #include <bits/stdc++.h>
 #define dbg(x) cout << #x << "=" << x << endl
 #define eps 1e-8
 #define pi acos(-1.0)

 using namespace std;
 typedef long long LL;

 template<class T>inline void read(T &res)
 {
     char c;T flag=;
     while((c=getchar())<''||c>'')if(c=='-')flag=-;res=c-'';
     while((c=getchar())>=''&&c<='')res=res*+c-'';res*=flag;
 }

 namespace _buff {
     const size_t BUFF =  << ;
     char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
     char getc() {
         if (ib == ie) {
             ib = ibuf;
             ie = ibuf + fread(ibuf, , BUFF, stdin);
         }
         return ib == ie ? - : *ib++;
     }
 }

 int qread() {
     using namespace _buff;
     int ret = ;
     bool pos = true;
     char c = getc();
     for (; (c < '' || c > '') && c != '-'; c = getc()) {
         assert(~c);
     }
     if (c == '-') {
         pos = false;
         c = getc();
     }
     for (; c >= '' && c <= ''; c = getc()) {
         ret = (ret << ) + (ret << ) + (c ^ );
     }
     return pos ? ret : -ret;
 }

 const int maxn = 5e4 + ;

 int fa[maxn];
 int tohead[maxn], sum[maxn];
 int t;

 int fid(int x) {
     if(fa[x] == x) {
         return x;
     }
     int r1 = fa[x], r2 = fid(fa[x]);
     fa[x] = r2;
     tohead[x] += tohead[r1];
     return r2;
 }

 void join(int x, int y) {
     int fx = fid(x), fy = fid(y);
     fa[fy] = fx;
     tohead[fy] = sum[fx];
     sum[fx] += sum[fy];
     sum[fy] = ;
 }

 int main()
 {
     scanf("%d",&t);
     for ( int i = ; i <= ; ++i ) {
         fa[i] = i;
         sum[i] = ;
     }
     while(t--) {
         char s[];
         int x, y;
         cin >> s;
         scanf("%d %d",&x, &y);
         if(s[] == 'M') {
             join(x, y);
         }
         else {
             if(fid(x) != fid(y)) {
                 printf("-1\n");
                 continue;
             }
             else {
                 printf("%d\n",abs(tohead[x] - tohead[y]) - );
             }
         }
     }
     return ;
 }